Interactive Lecture
Gear Ratios, Epicyclic Gear Trains, and Power Transmission
by Dr Arnaldo Delli Carri
Gear trains are mechanical systems composed of two or more gears that mesh together to transmit rotational motion and power. They are fundamental components in mechanical engineering, used to:
The gear ratio ties output speed and torque to the input. Gear down and you trade speed for torque; gear up and you trade torque for speed.
Gear Ratio: \(\eta = \frac{\omega_{driver}}{\omega_{driven}} = \frac{T_{driven}}{T_{driver}} = \frac{z_{driven}}{z_{driver}}\)
Where:
The module is the fundamental parameter in metric gear design. ISO 54:1996 specifies preferred modules:
Series 1 (Preferred): 0.5, 0.8, 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 16, 20, 25, 32, 40, 50 mm
Series 2: 0.55, 0.7, 0.9, 1.125, 1.375, 1.75, 2.25, 2.75, 3.5, 4.5, 5.5, 7, 9, 11, 14, 18, 22, 28, 36, 45 mm
Pitch Diameter: \(d = m \times z\)
Base Circle Diameter: \(d_b = d \times \cos(\alpha)\)
Addendum: \(h_a = 1.0 \times m\)
Dedendum: \(h_f = 1.25 \times m\)
Clearance: \(c = 0.25 \times m\)
Outside Diameter: \(d_a = d + 2h_a = m(z + 2)\)
Root Diameter: \(d_f = d - 2h_f = m(z - 2.5)\)
Where: \(m\) = module, \(z\) = number of teeth, \(\alpha\) = pressure angle
The involute curve is the standard profile for gear teeth. Watch how the curve is generated as if a string unwraps from a base circle. This simulation follows ISO standards for gear geometry.
The involute profile keeps the velocity ratio constant as the teeth mesh, which is what makes power transmission smooth. Contact between the teeth travels along a straight line, the line of action.
In a simple gear train, every gear rides on its own shaft. Drop an idler between the driver and driven gears and you reverse the output's direction without touching the gear ratio.
Fix two gears to a shared shaft and you have a compound train. Stacking stages this way reaches large ratios in a compact space.
For compound gear trains:
\(\eta_{1n} = \eta_{12} \times \eta_{23}\times \ldots \times\eta_{n-1,n}\)
An epicyclic (planetary) train nests a sun gear, orbiting planet gears carried on an arm, and an outer ring gear. Hold one member still and the same hardware delivers a different ratio.
Willis Equation for Epicyclic Gears:
\(\frac{\omega_s - \omega_c}{\omega_r - \omega_c} = -\frac{z_r}{z_s}\)
Where: s = sun, r = ring, c = carrier
Gear trains are ubiquitous in mechanical power transmission systems:
| Concept | Equation | Description |
|---|---|---|
| Gear Ratio | \(\eta = \frac{\omega_{in}}{\omega_{out}}=\frac{T_{out}}{T_{in}}=\frac{z_{out}}{z_{in}}\) | Speed ratio between gears |
| Pitch Diameter | \(d = mz\) | Diameter at tooth pitch circle |
| Willis Equation | \(\frac{\omega_s - \omega_c}{\omega_r - \omega_c} = -\frac{z_r}{z_s}\) | For Epicyclic Gear Trains |
Design a gear pair with a ratio of 3:1 using ISO standard modules. The pinion should have 20 teeth with module 4 mm and 20° pressure angle. Calculate:
Show Solution
1. Wheel specifications
A 3:1 ratio means the wheel needs three times the pinion's teeth: \(z_2 = 3 \times 20 = 60\) teeth. Meshing gears share the same module and pressure angle, so \(m = 4\,\text{mm}\) and \(\alpha = 20°\).
2. Centre distance
\[d_1 = m z_1 = 4 \times 20 = 80\,\text{mm}, \quad d_2 = m z_2 = 4 \times 60 = 240\,\text{mm}\]
\[a = \frac{d_1 + d_2}{2} = \frac{80 + 240}{2} = 160\,\text{mm}\]
3. Base circle diameters
\[d_{b1} = d_1 \cos\alpha = 80 \cos 20° = 75.18\,\text{mm}\]
\[d_{b2} = d_2 \cos\alpha = 240 \cos 20° = 225.5\,\text{mm}\]
Answer: wheel of 60 teeth (\(m = 4\,\text{mm}\), \(\alpha = 20°\)); centre distance \(a = 160\,\text{mm}\); base circles \(d_{b1} = 75.18\,\text{mm}\), \(d_{b2} = 225.5\,\text{mm}\).
A gear needs a pitch diameter of approximately 100 mm with 25 teeth. Select the appropriate ISO standard module and calculate the exact pitch diameter.
Show Solution
Rearranging \(d = mz\) for the module:
\[m = \frac{d}{z} = \frac{100}{25} = 4.0\,\text{mm}\]
This lands exactly on the ISO Series 1 preferred module of 4.0 mm, so the pitch diameter is met precisely:
\[d = m z = 4 \times 25 = 100\,\text{mm}\]
\[d_a = m(z + 2) = 4(25 + 2) = 108\,\text{mm}, \quad d_f = m(z - 2.5) = 4(25 - 2.5) = 90\,\text{mm}\]
Answer: module \(m = 4.0\,\text{mm}\), giving \(d = 100\,\text{mm}\), \(d_a = 108\,\text{mm}\), \(d_f = 90\,\text{mm}\).
A compound gear train consists of four gears arranged as follows:
Calculate:
Show Solution
(a) Overall gear ratio:
For a compound gear train, the overall ratio is the product of individual ratios:
\[\eta_{total} = \frac{z_2}{z_1} \times \frac{z_4}{z_3} = \frac{60}{20} \times \frac{45}{15} = 3 \times 3 = 9\]
The overall gear ratio is 9:1 (reduction)
(b) Output speed:
\[\omega_4 = \frac{\omega_1}{\eta_{total}} = \frac{1500}{9} = 166.7\,\text{RPM}\]
(c) Output torque:
For 100% efficiency, power is conserved: \(P_{in} = P_{out}\)
\[T_1 \omega_1 = T_4 \omega_4\]
\[T_4 = T_1 \times \frac{\omega_1}{\omega_4} = T_1 \times \eta_{total} = 50 \times 9 = 450\,\text{N·m}\]
Compound gear trains multiply torque by the overall ratio. This is why gearboxes convert high-speed, low-torque inputs into low-speed, high-torque outputs for heavy loads.
Answer: (a) \(\eta_{total} = 9\) (9:1 reduction); (b) \(\omega_4 = 166.7\,\text{RPM}\); (c) \(T_4 = 450\,\text{N·m}\).
An epicyclic gear train has a sun gear with 20 teeth and planet gears with 10 teeth each. The ring gear is held fixed and the sun is driven at 30 RPM. Calculate:
Show Solution
(a) Ring teeth: the planets bridge the sun and ring, so
\[z_r = z_s + 2 z_p = 20 + 2(10) = 40 \text{ teeth}\]
(b) Carrier speed: apply the Willis equation with the ring fixed, \(\omega_r = 0\):
\[\frac{\omega_s - \omega_c}{\omega_r - \omega_c} = -\frac{z_r}{z_s} \;\Rightarrow\; \frac{\omega_s - \omega_c}{-\omega_c} = -\frac{40}{20} = -2\]
\[\omega_s - \omega_c = 2\omega_c \;\Rightarrow\; \omega_s = 3\omega_c \;\Rightarrow\; \omega_c = \frac{30}{3} = 10\,\text{RPM}\]
(c) Gear ratio: with the ring fixed the sun drives the carrier, so
\[\eta = \frac{\omega_s}{\omega_c} = 1 + \frac{z_r}{z_s} = 1 + \frac{40}{20} = 3\]
The ratio is positive, so the carrier turns the same way as the sun.
Answer: (a) \(z_r = 40\) teeth; (b) \(\omega_c = 10\,\text{RPM}\); (c) \(\eta = 3\) (3:1 reduction), carrier same direction as sun.